That should work. I had a different answer in mind.
Spoiler forEncode days of the week with numbers: Sun=1, Mon=2, etc.
Whoever starts off with the mice sends a mouse clockwise N times, where N is the day of his execution, and each of the other prisoners will let the mouse continue on its way so all prisoners will see a mouse going clockwise N times. The starting prisoner then sends a mouse counterclockwise, with everyone passing the mouse along, to let everyone know they've reached the end of the number.
To get the mice to the next prisoner, the prisoner who started off with the mice sends one mouse clockwise and one mouse counterclockwise. The next prisoner (going clockwise) will see a mouse going clockwise after the counterclockwise mouse signaled the end of a number, and that will be his cue to keep the mice and become the next prisoner to signal his execution day. The other prisoners will see the mouse going counterclockwise (which will be the second time in a row they see one going counterclockwise), and will know that it's not their turn to signal yet and will let the counterclockwise mouse continue along until it reaches the next person to signal their date.
Then just repeat that process for all the prisoners. It should work as long as the distances between cells aren't vastly different -- if they were, then on the last step where the mice are released at the same time going in opposite directions, it might cause the counterclockwise mouse to arrive at the next cell going clockwise before the clockwise mouse.
Spoiler for In that case...A slightly more efficient method:
Assuming the mice move at the same speed (since they are alike in every other way and move at fixed speed), there are four possible easily distinguishable configurations:
A)1 mouse clockwise
B)1 mouse counterclockwise
C)2 mice clockwise
S)2 mice counterclockwise
Denote one of the last two as the 'STOP' code, let's say 2 mice counterclockwise. Agree that the order of signaling (i.e. the way the mice will be sent after a prisoner is done signalling his day of execution) will be the same direction, i.e. here counterclockwise.
You should be able to code all 7 days with 2 trips of the mice around, since there are 3*3=9 different combinations, as long as each time the prisoner allows the mice to return to his cell before sending the next trip (the time between trips that the other prisoners see may be significantly different, but that should be okay, since it's the order that matters, not the time). For example:
AA - Mon
AB - Tue
AC - Wed
BA - Thur
BB - Fri
BC - Sat
CC - Sun
After coding his day, the first prisoner who has the mice will send them both counterclockwise (S-stop) to allow everyone to know he's done. The first time each prisoner sees the stop code he will let the mice pass. The second time a prisoner sees this, he will know it's his turn and pick up the mice and start his code, so that the prisoner adjacent counterclockwise to the previous prisoner will be the one to pick up the mice and start his coding.
Edit: Actually, to make it the tiniest bit more efficient, since we're using a STOP code, you could code 3 of the days with just one trip, then STOP. Eh.